Kepler's Third Law Calculator
Kepler's third law links the orbital period of a planet or satellite to the size of its orbit and the mass of the body it circles. Enter any two of the three quantities and this calculator solves for the third instantly. Choose what you want to find from the drop-down, fill in the two known values, and read off the result with a full worked-step breakdown.
Formula
Worked example
Earth orbits the Sun at a = 1 AU with M = 1 solar mass. Plug in: T^2 = (4*pi^2 / (G * M_sun)) * (1 AU)^3. In natural units (AU, years, solar masses) this simplifies to T^2 = a^3, giving T = 1 year. For Mars (a = 1.524 AU): T = sqrt(1.524^3) = sqrt(3.540) = 1.881 years, matching the observed value of 687 days.
What is Kepler's third law?
Kepler's third law, published in 1619, states that the square of a planet's orbital period is proportional to the cube of its mean distance from the Sun. Johannes Kepler deduced this empirically from Tycho Brahe's meticulous planetary observations, long before Newton derived it from the law of universal gravitation. In modern form, combining Newton's gravitational law with circular-orbit dynamics gives T^2 = (4*pi^2 / G*M) * a^3, where T is the orbital period, a is the semi-major axis, G is the gravitational constant (6.674 x 10^-11 m^3 kg^-1 s^-2), and M is the mass of the central body. The elegant result is that for any body orbiting the same star, the ratio a^3/T^2 is the same constant.
How this calculator works - three solve modes
Select what you want to find from the 'Solve for' dropdown. If you want the orbital period (T), enter the semi-major axis in AU and the central body mass in solar masses: the calculator evaluates T = sqrt(4*pi^2 * a^3 / G*M). If you want the semi-major axis (a), enter the orbital period and central body mass: the calculator inverts the formula to a = cbrt(G*M*T^2 / 4*pi^2). If you want the central body mass (M), enter the orbital period and semi-major axis: M = 4*pi^2 * a^3 / (G*T^2). All three directions are computed in SI (metres, kilograms, seconds) then converted to the display units (AU, years, solar masses).
Units and how to convert them
This calculator uses astronomical units (AU) for distance, Earth years for time, and solar masses for mass. One AU is the mean Earth-Sun distance, 149,597,870.7 km. One solar mass is 1.989 x 10^30 kg. The law applies universally: to calculate a satellite orbit around Earth, express Earth's mass in solar masses (3.003 x 10^-6) and the orbit radius in AU. For very tight orbits (artificial satellites), the period comes out in fractions of a year, which the calculator converts to hours or days automatically. The "a^3/T^2" output is Kepler's constant for the system, numerically equal to the central body mass in solar masses when AU and years are used.
Historical significance and modern applications
Kepler's third law was the first quantitative relationship connecting two bodies that are not in direct contact - a conceptual leap that helped Newton formulate universal gravitation. Today the same formula is used to determine the mass of distant stars from the periods and orbits of their planets (exoplanet science), to design stable satellite orbits, to calculate the orbital parameters of binary star systems, and to estimate the mass enclosed within a galaxy from the speeds of its outermost stars. The law assumes a two-body system with one mass much larger than the other. For binary stars of comparable mass the full form T^2 = 4*pi^2 * a^3 / (G*(M+m)) is needed, where m is the orbiting body's mass.
Solar system planetary data
| Body | Semi-major axis (AU) | Period (years) | a³/T² (AU³/yr²) |
|---|---|---|---|
| Mercury | 0.387 | 0.241 | 0.9979 |
| Venus | 0.723 | 0.615 | 0.9992 |
| Earth | 1 | 1 | 1.0000 |
| Mars | 1.524 | 1.881 | 1.0004 |
| Jupiter | 5.203 | 11.862 | 1.0010 |
| Saturn | 9.537 | 29.457 | 0.9997 |
| Uranus | 19.191 | 84.021 | 1.0012 |
| Neptune | 30.069 | 164.8 | 1.0010 |
| Pluto | 39.482 | 247.94 | 1.0012 |
Orbital periods and semi-major axes for the eight planets and Pluto. Notice that a^3/T^2 is nearly constant across all bodies, demonstrating Kepler's third law.
Frequently asked questions
What does Kepler's third law say?
Kepler's third law states that the square of the orbital period (T^2) is proportional to the cube of the semi-major axis of the orbit (a^3). In equation form: T^2 / a^3 = constant for all bodies orbiting the same central mass. When the units are years and AU, the constant equals 1 for our solar system (where M = 1 solar mass), so T^2 = a^3 directly.
What are the inputs and outputs of this calculator?
The calculator works with three quantities: orbital period (T) in years, semi-major axis (a) in astronomical units, and central body mass (M) in solar masses. Select which one you want to find, enter the other two, and the calculator returns the missing value along with conversions to days, kilometres and kilograms, the mean orbital speed, and Kepler's constant a^3/T^2 for verification.
Why is the constant a^3/T^2 close to 1 for solar-system planets?
When distances are measured in AU and time in Earth years, the constant G*M/(4*pi^2) evaluates to 1 AU^3/yr^2 for M = 1 solar mass. Because all eight planets orbit the same Sun, a^3/T^2 should be approximately 1 for every planet. You can verify this in the planet reference table: Mercury gives (0.387^3)/(0.241^2) = 0.0580/0.0581 = 0.9983, Earth gives exactly 1, and so on. Small deviations are due to rounding the tabulated values.
Can I use this calculator for artificial satellites?
Yes. Use Earth's mass in solar masses (about 3.003 x 10^-6) and the orbit radius in AU (1 AU = 149,597,870.7 km, so a low-Earth orbit at ~400 km altitude is about (6,771 km / 149,597,871 km/AU) = 4.526 x 10^-5 AU). The calculator will give a period in fractions of a year, which it converts automatically to hours or days. The International Space Station's ~92-minute period corresponds to a semi-major axis of roughly 4.5 x 10^-5 AU.
Does the formula work for binary star systems?
Not directly in its simplified form. For a binary star where both masses (M and m) are significant, the correct form is T^2 = 4*pi^2 * a^3 / (G*(M+m)), where a is the separation between the two stars. If you substitute M+m for M in this calculator, you can still solve for the orbital separation or period correctly.
What is the semi-major axis and how is it related to the mean distance?
The semi-major axis is half the longest diameter of an elliptical orbit. For a circle it equals the radius. Kepler's third law uses the semi-major axis rather than the average separation because it is the quantity that naturally appears when you integrate the orbital energy. For orbits with low eccentricity (like Earth's, e = 0.017), the semi-major axis is nearly equal to the average Sun-planet distance, which is why the terms are often used interchangeably in introductory courses.